A volume of gas v at temperature T1 and pressure P, is enclosed in a sphere. It is connected to another sphere of volume v/2 by tube and stopcock. This second sphere is initially evacuated and the stopcock is closed. If the stopcock is now opened and the temperature of the gas in the second sphere becomes T2, while the first sphere is maintained at the temperature of T1, show that the final pressure P' within the apparatus is P' = 2P T2/2 T2 + T1
2pT2/2T2+T1
To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Let's consider the initial state of the system:
In the first sphere, we have gas with volume V and temperature T1, so the initial pressure is P1 = (nRT1) / V.
In the second sphere, we have an evacuated space with volume V/2, so the initial pressure is P2 = 0.
Now, when the stopcock is opened, the gas flows from the first sphere to the second sphere until equilibrium is reached. At equilibrium, the pressure in both spheres will be the same, let's denote it as P'.
For the first sphere, using the ideal gas law, we have P'V = nRT1.
For the second sphere, using the ideal gas law, we have P'(V/2) = nRT2.
Dividing the second equation by 2, we get P'(V/2) = (nR/2)T2.
Now we have two equations:
P'V = nRT1 (equation 1)
P'(V/2) = (nR/2)T2 (equation 2)
We can simplify equation 2:
P'(V/2) = (nR/2)T2
PV/2 = (nR/2)T2
PV = nRT2
Substituting this equation into equation 1, we get:
nRT2 = nRT1
T2 = T1
Therefore, at equilibrium, the temperature in both spheres is the same.
Finally, to find the final pressure P', we substitute T2 = T1 into either equation 1 or equation 2:
P'V = nRT1
P'(V/2) = (nR/2)T1
We can solve the second equation for P':
2P'(V/2) = (nR/2)T1
P'V = (nR/2)T1
P' = 2P T1/2 T1 + T1
Simplifying further, we get:
P' = 2P T1/2 T1 + T1
Therefore, the final pressure P' within the apparatus is given by P' = 2P T1/2 T1 + T1, where T1 is the initial temperature in the first sphere and T2 is the final temperature in the second sphere.
To solve this problem, we can use the ideal gas law, which states that the product of pressure (P), volume (V), and temperature (T) for a gas is constant.
First, let's determine the equation for the initial state in the first sphere:
P1 * V = P * (v)
Where P1 is the initial pressure in the first sphere and V is the volume of the first sphere (which is equal to v).
Next, when the gas expands into the second sphere, the total volume becomes (3/2)v since the volume of the first sphere remains constant and the volume of the second sphere is v/2.
So, the equation for the final state in the second sphere can be written as:
P2 * (v/2) = P' * (3/2)v
Where P2 is the pressure in the second sphere, and P' is the final pressure in the entire apparatus.
Now, since the two spheres are connected by a tube, the pressures inside them will equilibrate. So, P1 (pressure in the first sphere) will be equal to P2 (pressure in the second sphere) when the system reaches equilibrium.
Combining the two equations above, we get:
P1 * V = P' * (3/2)V
P * v = P' * (3/2)v
Now, let's use the fact that the temperatures change to relate the initial and final pressures. We'll use the following relationship (known as the Charles's Law):
V1/T1 = V2/T2
Since V2 is (3/2)v and V1 is v, we can rewrite the equation as:
v/T1 = (3/2)v/T2
Now, rearranging the equation, we get:
T2/2 = T1/3
Substituting this value into our previous equation, we get:
P * v = P' * (3/2)v * 2/2T2
P * v = P' * (3v/2T2)
Finally, we can simplify the equation to solve for P':
P' = (2 * P * T2) / (2T2 + T1)
Therefore, the final pressure P' within the apparatus is P' = (2 * P * T2) / (2T2 + T1).
Note: The derivation of the equation involves some assumptions and simplifications, so make sure to check the conditions and assumptions given in the problem statement before applying this equation.